Comparing Kaplan-Meier and Exponentiated Nelson-Aalen Estimates

Why Compare the Two Survival Estimates?

The Kaplan-Meier and Nelson-Aalen methods use the same observed failures, censoring information, and risk sets, but summarize them differently. Kaplan-Meier estimates survival by multiplying conditional survival factors. Nelson-Aalen estimates cumulative hazard by adding hazard increments; exponentiating the negative of that total gives another estimate of survival. The two results are related, but with tied failures they need not be equal.

The comparison matters because the exponentiated Nelson-Aalen estimate is sometimes used as an approximation to Kaplan-Meier survival. For small event fractions at each event time, the two can be close. If an appreciable fraction of a risk set fails at one time, their difference can be more noticeable. This tutorial is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: At distinct observed failure time \(t_j\), let \(n_j\) be the number at risk immediately before \(t_j\), and let \(d_j\) be the number of failures tied at \(t_j\). The Kaplan-Meier estimate through time \(t\) is the product of its conditional survival factors over failure times \(t_j \leq t\). The Nelson-Aalen cumulative hazard estimate adds the increments \(d_j/n_j\). Its exponentiated survival estimate is the exponential of the negative cumulative hazard estimate.

\widehat S_{\mathrm{KM}}(t) = \prod_{t_j \leq t}\left(1-\frac{d_j}{n_j}\right), \qquad \widehat H_{\mathrm{NA}}(t) = \sum_{t_j \leq t}\frac{d_j}{n_j}, \qquad \widehat S_{\mathrm{NA,exp}}(t) = \exp\left\{-\widehat H_{\mathrm{NA}}(t)\right\}.

These are nonparametric estimates: they use the event times, censoring records, and risk sets, not a specified parametric lifetime distribution. At a tied failure time, the whole group of \(d_j\) failures contributes one factor to Kaplan-Meier and one increment to Nelson-Aalen. Under the usual risk-set convention, people censored at a failure time are included in the risk set immediately before that time and are removed afterward; they do not count as failures.

Why the Estimates Differ

Write \(q_j=d_j/n_j\), the fraction of the risk set that fails at time \(t_j\). The Kaplan-Meier factor at that time is \(1-q_j\); the corresponding exponentiated Nelson-Aalen factor is \(e^{-q_j}\). For \(0 \leq q_j < 1\), these factors are not the same. The logarithm identity below explains both the approximation and the direction of the difference.

Express the Kaplan-Meier estimate on the log scale.
Taking logarithms of its product gives \(\log\widehat S_{\mathrm{KM}}(t)=\sum_{t_j\leq t}\log(1-q_j)\).

Compare one event-time contribution.
For \(0 \leq q_j < 1\), \(\log(1-q_j)=-q_j-q_j^2/2-q_j^3/3-\cdots \leq -q_j\).

Add the contributions and exponentiate.
Summing gives \(\log\widehat S_{\mathrm{KM}}(t)\leq-\sum q_j=-\widehat H_{\mathrm{NA}}(t)\). Since the exponential function is increasing, \(\widehat S_{\mathrm{KM}}(t)\leq\widehat S_{\mathrm{NA,exp}}(t)\).

The series also shows why the estimates can be close: when each \(q_j\) is small, the leading difference between \(\log(1-q_j)\) and \(-q_j\) is approximately \(-q_j^2/2\). But the discrepancy accumulates over event times. If \(q_j=1\), all individuals at risk fail at that time: Kaplan-Meier becomes zero, whereas the exponentiated Nelson-Aalen factor is \(e^{-1}\), which is positive. The inequality still holds, although the logarithmic series argument is applied by taking the limit as \(q_j\) approaches one.

Key result: At a common time point, the exponentiated Nelson-Aalen estimate is at least as large as the Kaplan-Meier estimate. Equality holds when there have been no failures, and otherwise the event-time factors make the estimates different. “At least as large” describes the calculated estimates; it does not establish which one is closer to the unknown true survival probability.

Worked Comparisons

Worked Example: Several Tied-Failure Times

A hypothetical group is followed for time to equipment failure. At time 2, two units fail from a risk set of 20, and three other units are censored after that event time. At time 4, four fail from the resulting risk set, and two are censored after the failures. At time 7, three fail. Compare the estimates through time 7. This is a nonparametric calculation: no interest rate, mortality law, or parametric lifetime distribution is assumed.

Step 1: Establish the risk sets. The initial risk set is 20. After two failures and three censorings, the risk set at time 4 is \(20-2-3=15\). After four failures and two censorings, the risk set at time 7 is \(15-4-2=9\). Thus \((n_j,d_j)\) is \((20,2)\), \((15,4)\), and \((9,3)\).

Step 2: Calculate Kaplan-Meier survival. The three conditional survival factors are \(1-2/20=18/20=0.9\), \(1-4/15=11/15\), and \(1-3/9=6/9=2/3\). Therefore,

\widehat S_{\mathrm{KM}}(7) = \frac{18}{20}\cdot\frac{11}{15}\cdot\frac{6}{9} = \frac{9}{10}\cdot\frac{11}{15}\cdot\frac{2}{3} = \frac{11}{25} = 0.4400.

Step 3: Calculate and exponentiate the Nelson-Aalen estimate. The hazard increments are \(2/20=0.1000\), \(4/15=0.2667\), and \(3/9=0.3333\). Their sum is \(0.1000+0.2667+0.3333=0.7000\), so \(\widehat S_{\mathrm{NA,exp}}(7)=e^{-0.7}=0.4966\), rounded to four decimal places.

Step 4: Compare and check. The difference, using the unrounded exponential value, is \(0.4965853-0.4400=0.0565853\), or \(0.0566\) to four decimals. The exponentiated Nelson-Aalen estimate is larger, as the inequality predicts. The substantial event fractions at times 4 and 7 help explain the gap.

Method and trap avoided: Both estimates use the same risk sets, but one multiplies \(1-d_j/n_j\) while the other exponentiates the sum of \(d_j/n_j\). Treating those operations as identical would incorrectly report \(0.4400\) for both.

Worked Example: Small Event Fractions

In a hypothetical reliability study, two units fail from a risk set of 100 at the first event time. At the next event time, three fail from the remaining risk set. Find both survival estimates through the second event time and compare them. No interest rate, mortality law, or parametric lifetime distribution is assumed.

Step 1: Determine the second risk set. There are \(100-2=98\) units at risk at the second event time, assuming no intervening censoring.

Step 2: Calculate Kaplan-Meier survival. The factors are \(1-2/100=0.98\) and \(1-3/98=95/98\). Their product is \(0.98(95/98)=(98/100)(95/98)=95/100=0.9500\).

Step 3: Calculate exponentiated Nelson-Aalen survival. The cumulative hazard estimate is \(2/100+3/98=0.020000+0.0306122=0.0506122\). Hence the exponentiated estimate is \(e^{-0.0506122}=0.9506\), rounded to four decimal places.

Step 4: Interpret. Using the unrounded exponential value, the gap is approximately \(0.950647-0.950000=0.000647\), or \(0.0006\) to four decimals. The event fractions are small, so the two estimates are close, but not equal.

Method and trap avoided: The second event fraction uses 98, not 100, because two units already failed. Reusing the original risk set would give an incorrect second increment and an incorrect comparison.

Worked Example: A Large Tied-Failure Fraction

A hypothetical cohort has 12 units at risk at one observed failure time, and six fail together. Compare the two estimates immediately after that time. No interest rate, mortality law, or parametric lifetime distribution is assumed.

Step 1: Calculate Kaplan-Meier survival. Here \(q=6/12=0.5\), so the Kaplan-Meier factor is \(1-q=0.5\). Thus \(\widehat S_{\mathrm{KM}}=0.5000\).

Step 2: Calculate exponentiated Nelson-Aalen survival. The cumulative hazard increment is \(6/12=0.5\). Therefore, \(\widehat S_{\mathrm{NA,exp}}=e^{-0.5}=0.6065\), rounded to four decimal places.

Step 3: Compare. The difference is \(0.6065307-0.5000=0.1065307\), or \(0.1065\) to four decimals. This larger gap, compared with the previous example, is consistent with the larger event fraction.

Method and trap avoided: A tied group of six failures contributes \(6/12\) to the Nelson-Aalen increment and a single factor \(1-6/12\) to Kaplan-Meier. It is not six sequential failures with six changing risk sets: the failures are tied at one recorded time.

Interpreting the Difference Correctly

A useful comparison should report the time point, both estimates, and their difference. For example, in the first worked example, estimated survival through time 7 is \(0.4400\) by Kaplan-Meier and \(0.4966\) by the exponentiated Nelson-Aalen method, a difference of \(0.0566\). Stating only that one method “approximates” the other hides how material the discrepancy is in the particular calculation.

Censoring matters through the risk sets: it changes the denominators \(n_j\) at later event times. A censoring observation does not itself add a Kaplan-Meier survival factor or a Nelson-Aalen hazard increment. A tied failure group, by contrast, contributes using the risk set just before its event time. Keeping these roles separate prevents both incorrect denominators and incorrect extra factors.

The gap is not a correction that should automatically be added to either estimate. The two procedures are different summaries of the observed event history. The exponential transformation is motivated by the relationship between cumulative hazard and survival, but for finite jumps its exponential form does not reproduce the Kaplan-Meier product. The practical decision is to calculate the requested estimate precisely and, when comparing methods, interpret the size and direction of the difference rather than silently substituting one for the other.

Common trap: Replacing every Kaplan-Meier factor \(1-d_j/n_j\) with \(e^{-d_j/n_j}\) and then calling the result the Kaplan-Meier estimate. The replacement gives the exponentiated Nelson-Aalen estimate. Since \(e^{-q_j}\geq 1-q_j\), the resulting survival estimate is at least as large, and can differ appreciably when event fractions are large.

Check Your Understanding

For each problem, calculate the Kaplan-Meier estimate and the exponentiated Nelson-Aalen estimate through the stated event time or times. Round estimates and differences to four decimal places.

A hypothetical cohort has 40 units at risk at one event time, with four tied failures. Find both estimates and their difference. (Answer: Kaplan-Meier \(0.9000\); exponentiated Nelson-Aalen \(0.9048\); difference \(0.0048\).)
At the first event time, five of 25 units fail. At the second, two of the remaining 20 fail, with no censoring in between. Find both estimates through the second event time and their difference. (Answer: Kaplan-Meier \(0.7200\); exponentiated Nelson-Aalen \(0.7408\); difference \(0.0208\).)
Ten of 50 units fail at the first event time; then eight of the remaining 40 fail at the next. Find both estimates and their difference. (Answer: Kaplan-Meier \(0.6400\); exponentiated Nelson-Aalen \(0.6703\); difference \(0.0303\).)
The risk sets and tied failures at three times are \((18,3)\), \((12,3)\), and \((8,4)\). Calculate both estimates through the third time and their difference. (Answer: Kaplan-Meier \(0.3125\); exponentiated Nelson-Aalen \(0.3998\); difference \(0.0873\).)
A hypothetical cohort starts with 30 units. At time 1, three fail and six are censored after the event time. At time 3, seven fail and two are censored after the event time. At time 5, four fail. Find both estimates through time 5 and their difference. (Answer: Kaplan-Meier \(0.4000\); exponentiated Nelson-Aalen \(0.4646\); difference \(0.0646\).)

What You'll Learn